Finding the Limit of Binomial CDF

Question

Let $X\sim \mathrm{Binom}(n,p)$. Then its CDF is $$F_X(x)=\sum_{k=0}^{\lfloor x\rfloor}\binom{n}{k}p^k(1-p)^{n-k}.$$ I am trying to determine the limit of $F_X(x)$ with respect to $n$, with $p$ remain fixed: $$\lim_{n\to\infty}F_X(x)=\lim_{n\to\infty}\sum_{k=0}^{\lfloor x\rfloor}\binom{n}{k}p^k(1-p)^{n-k}.$$ The first thing that came up is de Moivre–Laplace theorem, that is $\frac{X-np}{\sqrt{np(1-p)}}\xrightarrow{D}N(0,1)$. Can I conclude from that fact then "$X\xrightarrow{D} N(np, np(1-p))$" (not really since $np$ explodes), so $$\lim_{n\to\infty}F_X(x)=\lim_{n\to\infty}\left[\frac{1}{2}+\frac{1}{2}\mathscr{E}\mathrm{rf}\left(\frac{x-np}{\sqrt{2np(1-p)}}\right)\right]?$$ If not, how should I continue?

Answer 1

The de Moivre–Laplace limit theorem involves relocation by subtraction of $np$ and rescaling by dividing by $\sqrt{np(1-p)}$. If you just want the limit of the c.d.f. at a fixed value of $x$ without doing those things, then the limit is $0$. For example, suppose $x=40$ and $p=0.2$ and $n=10000000000000000000$. What would you expect the probability that $X\le x$ to be?

PS: Here's a simple way to prove the limit is $0$. As to the rate of convergence, this method will be conservative.

Chebyshev's inequality says that if a random variable $Y$ has expected value $\mu$ and variance $\tau^2$ then $$ \Pr\left( |Y-\mu| > k \right) \le \frac{\tau^2}{k^2}. $$

You have $\operatorname{E}(X)=np$ and $\operatorname{var}(X)= np(1-p)$ so \begin{align} \Pr(X=x) \le \Pr(X < x+1) &= \Pr\left( X-np < x+1-np \right) \le \Pr\left( |X-np|> |x+1-np| \right) \\[10pt] & \le \frac{np(1-p)}{(x+1-np)^2} \to 0 \text{ as } n\to\infty \end{align} since we have the square of $n$ in the denominator and the first power of $n$ in the numerator.

Answer 2

You are on the right track.

A naive search for "normal approximation to the binomial distribution" produces many useful links.